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APM346-2018S => APM346--Tests => Term Test 2 => Topic started by: Victor Ivrii on March 23, 2018, 06:10:04 AM

Title: TT2--P3
Post by: Victor Ivrii on March 23, 2018, 06:10:04 AM

Using Fourier method find eigenvalues and eigenfunctions of Laplacian in the rectangle $\{0<x<a,\, 0<y<b\}$ with the boundary conditions:
\begin{align}
&u_{xx}+u_{yy}=-\lambda u\qquad 0<x<a,\ 0<y<b,\label{3-1}\\[3pt]
&u_x|_{x=0}=u_x|_{x=a}=u|_{y=0}=u|_{y=b}=0.\label{3-2}
\end{align}

Title: Re: TT2--P3
Post by: Jilong Bi on March 23, 2018, 09:42:24 AM

First separate of variable $$u(x,y) =X(x)Y(y),$$
take derivative
$$ X''Y + XY'' = -\lambda XY$$
$$\implies  \frac{X''}{X}+\frac{Y''}{Y}= -\lambda $$
By the given condition
 $$ X''+\lambda_1X = 0 , X'(0)+X'(a) = 0 $$
This is Neumann boundary condition
As n = 0,$$ \lambda_1 = 0 ,X_0 = \frac{1}{2}$$
As n = 1,2,...,$$ \lambda_1 =  \frac{n^2\pi ^2}{a^2},X_1 = \cos\frac{n\pi x}{a}$$
For Y, this is Dirichlet boundary condition  $$ Y''+\lambda_2Y= 0 , Y(0)+Y(b) = 0 $$
As m = 1,2,...,$$ \lambda_2 =  \frac{m^2\pi ^2}{b^2},Y_1 = \sin\frac{m\pi y}{b}$$
$\lambda  =\lambda_1  + \lambda_2 $, for n,m = 1,2,....
$$\implies  \lambda =  \pi^2(\frac{m^2}{b^2}+\frac{n^2}{a^2}) $$
for n,m = 1,2,....
$$ u(x,y) =\cos\frac{n\pi x}{a} \sin\frac{m\pi y}{b}$$
For n = 0, $\lambda$  = 0 and u(x,y) = 0